Systems of Differential Equations
6.10 Applications
A. Introduction
In this section, we revisit the application of differential equations in modeling engineering systems. In particular, we focus on mechanical vibrations and electrical circuits as two primary areas where systems of differential equations are applied.
Differential equations have a broad utility across various engineering fields. In chemical engineering, they are pivotal for modeling reaction kinetics and process dynamics. This includes scenarios such as mixing problems involving multiple tanks and substances, which are essential for reactor design and process optimization. In civil engineering, differential equation models are crucial for assessing the safety and longevity of structures subjected to diverse load conditions, such as in the earthquake resilience analysis of multi-story buildings. Aerospace engineering relies on these equations to simulate the movement of aircraft and spacecraft, incorporating both translational and rotational dynamics. This knowledge is instrumental in crafting control systems that enhance stability and maneuverability. Environmental engineering also employs differential equation models to track pollutant spread, providing a foundation for crafting effective environmental protection measures.
B. Electrical Circuits
Kirchhoff’s laws, which we discussed in Section 2.5, serve as the foundation for deriving the governing equations. These laws facilitate the analysis of circuits by providing a systematic approach to calculating the currents and voltages at various points within the circuit. In more complex circuits, e.g., series-parallel circuits,
a) For the given electrical circuit diagram, derive the system of differential equations that describes the currents in various branches of the circuit. Assume that all initial currents are zero. b) Once the system of differential equations and initial conditions are established, solve the system for the currents in each branch of the circuit.
Diagram Description
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a) In Example 4.8.2, we previously examined this RL circuit and analyzed it using the Laplace Transform. In this example, we demonstrate solving the same circuit with the matrix method. The system equations for the circuit are given as follows, with initial conditions that all currents are zero at the time [asciimath]t=0[/asciimath].
[asciimath]{(4I_1+0.1I'_3+0.2I'_1=12),(0.1I'_3-2I_2=0), (I_1-I_2-I_3=0) :}[/asciimath] [asciimath];[/asciimath] [asciimath]I_1(0)=I_2(0)=I_3(0)=0[/asciimath] (6.10.1)
b) Steps for solving the system:
1. System 6.10.1 is a mix of differential and algebraic equations. We first need to convert it into a system of linear differential equations by using the second equation to express [asciimath]0.1I'_3[/asciimath] as [asciimath]2I_2[/asciimath] in the first equation and isolating the first derivatives in the first two equations. This yields
[asciimath]{(I'_1=-20I_1-10I_2+60),(I'_3=20I_2), (I_1-I_2-I_3=0) :}[/asciimath] [asciimath];[/asciimath] [asciimath]I_1(0)=I_2(0)=I_3(0)=0[/asciimath] (6.10.1)
To create a system of linear differential equations from the given system, it’s important to address the fact that [asciimath]I_2[/asciimath] does not have a derivative present. To work around this, [asciimath]I_2[/asciimath] needs to be eliminated from the equations. This is achieved by rearranging the third equation to express [asciimath]I_2[/asciimath] in terms of [asciimath]I_1[/asciimath] and [asciimath]I_3[/asciimath], and then substituting this expression for [asciimath]I_2[/asciimath] into the other equations.
[asciimath]I_2=I_1-I_3[/asciimath] (6.10.2)
The system is then simplified to
[asciimath]{(I'_1=-30I_1+10I_3+60),(I'_3=20I_1-20I_3) :}[/asciimath] [asciimath];[/asciimath] [asciimath]I_1(0)=I_3(0)=0[/asciimath] (6.10.2)
2. We then express the initial value problem (IVP) in matrix form.
[asciimath]bb(I)'=[(-30,10),(20,-20)] bb(I)+[(60),(0)],[/asciimath] [asciimath]bb(I)(0)=[(0),(0)][/asciimath]
3. Next, we find a fundamental solution to the associated complementary (homogeneous) system. The characteristic polynomial of [asciimath]A[/asciimath] is given by
[asciimath]c_A(lambda)=det(A-lambda I)[/asciimath]
[asciimath]=|(-30-lambda,10),(20,-20-lambda) |[/asciimath]
[asciimath]=(-30-lambda)(-20-lambda)-200[/asciimath]
[asciimath]=lambda^2+50lambda+400[/asciimath]
[asciimath]=(lambda+10)(lambda+40)[/asciimath]
The eigenvalues and the corresponding eigenvectors are
[asciimath]lambda_1=-10[/asciimath] : [asciimath]bb(u)_1=[(1),(2)][/asciimath]
[asciimath]lambda_2=-40[/asciimath] : [asciimath]bb(u)_1=[(-1),(1)][/asciimath]
Therefore, [asciimath]{e^(-10t) [(1),(2)], \ e^(-40t) [(-1),(1)] }[/asciimath] is a fundamental solution set to the complementary system.
Thus the fundamental matrix [asciimath]I_c(t)[/asciimath] for the complementary system is
[asciimath]I_c(t)=[(e^(-10t),-e^(-40t)),(2e^(-10t),e^(-40t))][/asciimath]
3. Next, we determine a particular solution to the system
(i) Determine [asciimath]I_c^-1(t)[/asciimath]
[asciimath]I_c^-1(t)=1/3[(e^(10t),e^(10t)),(-2e^(40t),e^(40t))][/asciimath]
(ii) Determine[asciimath]bb"v"(t)[/asciimath] letting the constant of integration be zero
[asciimath]bb"v"(t)=int\ I_c^-1(t)bb"f"(t) dt[/asciimath]
[asciimath]=int \ [(e^(10t),e^(10t)),(-2e^(40t),e^(40t))] [(60),(0)] dt[/asciimath]
[asciimath]=[(2e^(10t)),(-e^(40t))][/asciimath]
Then, a particular solution to the system is
[asciimath]bb"I"_p(t)=I_c(t).bb"v"(t)[/asciimath]
[asciimath]=[(e^(-10t),-e^(-40t)),(2e^(-10t),e^(-40t))] [(2e^(10t)),(-e^(40t))][/asciimath]
[asciimath]=[(3),(3)][/asciimath]
4. Thus, a general solution to the system is
[asciimath]bb"I"=bb(I)_p+bb(I)_c[/asciimath]
[asciimath]=[(3),(3)][/asciimath] [asciimath]+c_1e^(-10t) [(1),(2)] +c_2e^(-40t) [(-1),(1)][/asciimath]
5. We apply the initial conditions to find the constants in the general solution.
[asciimath]bb(I)(0)=[(0),(0)][/asciimath]
[asciimath][(3),(3)][/asciimath] [asciimath]+c_1e^(-10t) [(1),(2)] +c_2e^(-40t) [(-1),(1)]=[(0),(0)][/asciimath]
This gives a system of two equations and two unknowns.
[asciimath]{(c_1-c_2=-3),(2c_1+c_2=-3 ) :}[/asciimath]
Solving the system gives
[asciimath]c_1=-2,[/asciimath] [asciimath]c_2=1[/asciimath]
Therefore, the solution to the initial value problem is
[asciimath]bb(I)=[(3),(3)][/asciimath] [asciimath]-2e^(-10t) [(1),(2)] +e^(-40t) [(-1),(1)][/asciimath]
This results in the final expressions for [asciimath]I_1[/asciimath] and [asciimath]I_3[/asciimath].
[asciimath]I_1=3-2e^(-10t)-e^(-40t)[/asciimath]
[asciimath]I_3=3-4e^(-10t)+e^(-40t)[/asciimath]
6. To find [asciimath]I_2[/asciimath] , we substitute back the expression for [asciimath]I_1[/asciimath] and [asciimath]I_3[/asciimath] into Equation (6.10.1), yielding
[asciimath]I_2=I_1-I_3[/asciimath]
[asciimath]I_2=2e^(-10t)-2e^(-40t)[/asciimath]
C. Mechanical Vibration
The analysis of mechanical vibrations is crucial in designing systems that are resilient to dynamic loads. A more realistic model that captures the essence of mechanical systems involves considering not only the masses and springs but also damping elements and external forces. This section focuses on a system consisting of two masses connected by springs in a horizontal arrangement, with the inclusion of damping and external forces acting on both masses. Such a model can represent a wide array of engineering applications, from vehicle suspensions to machinery components. A schematic of the system is shown in Figure 6.10.1.
Assumptions
To proceed with the derivation of the system’s equations of motion, we make the following assumptions:
- Linear Damping: Each mass is paired with a damping element, characterized by linear damping coefficients [asciimath]c_1[/asciimath]for [asciimath]m_1[/asciimath] and [asciimath]c_2[/asciimath] for [asciimath]m_2[/asciimath]. These coefficients quantify the resistance against the motion of each mass.
- External Forces: Time-dependent external forces [asciimath]F_1(t)[/asciimath] and [asciimath]F_2(t)[/asciimath] act on [asciimath]m_1[/asciimath] and [asciimath]m_2[/asciimath], respectively, considered positive in the right direction.
- Linear Elasticity: The springs obey Hooke’s law, implying that the force each spring exerts is directly proportional to its displacement from the rest length.
- Small Displacements: The analysis assumes small displacements from equilibrium, allowing linearization of the system. Displacements are deemed positive when directed to the right.
- Rigid Body: Masses are treated as point masses, and springs and dampers are considered massless, focusing solely on axial forces and displacements.
System Setup
We consider a general case where the system consists of two masses, [asciimath]m_1[/asciimath] and [asciimath]m_2[/asciimath], connected by three springs with stiffness constants [asciimath]k_1[/asciimath], [asciimath]k_2[/asciimath], and [asciimath]k_3[/asciimath], and augmented by two dampers. The outer springs are anchored to fixed walls. External forces act upon the masses and dampers counteract their movement. This framework allows the external forces and damping effects to be adjustable, accommodating scenarios where these forces might be absent by setting their respective values to zero.
Derive the system of differential equations for the forced damped coupled system described above (Figure 6.10.1).
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The dynamics of this damped system with external forces are governed by two coupled second-order linear differential equations, reflecting the balance of forces on each mass. Here we consider the external forces’ direction to be to the right and displacements are also positive (to the right) assuming the displacement of mass 1, [asciimath]x_1[/asciimath] is larger than the displacement of mass 2, [asciimath]x_2[/asciimath], thus [asciimath]x_2-x_1<0[/asciimath].
1) The forces acting on mass [asciimath]m_1[/asciimath]are
- Restorative Force of the spring [asciimath]k_1[/asciimath] , [asciimath]F_(s1)=-k_1x_1[/asciimath],
- Restorative Force of the spring [asciimath]k_2[/asciimath] , [asciimath]F_(s2)=-k_2(x_2-x_1)[/asciimath], where [asciimath]x_2-x_1[/asciimath] is the displacement of the middle spring.
- Damping Force [asciimath]F_(d1)=-c_1x_1'[/asciimath], where [asciimath]c_1[/asciimath] is the damping coefficient for the damper 1. If present, the damping force is proportional to the velocity of the mass and acting in the opposite direction of motion.
- External Force [asciimath]F_1(t)[/asciimath]. It includes any external force acting on mass [asciimath]m_1[/asciimath], which might be periodic or random, leading to forced vibrations.
According to Newton’s second law,
[asciimath]ma=sumF[/asciimath]
[asciimath]m_1x''_1=-c_1x'_1-k_1x_1+k_2(x_2-x_1)+F_1(t)[/asciimath]
This equation simplifies to
[asciimath]m_1x''_1=-c_1x'_1-(k_1+k_2)x_1+k_2x_2+F_1(t)[/asciimath]
Note that since [asciimath]x_2ltx_1[/asciimath], the spring [asciimath]k_2[/asciimath] is compressed, and the force it exerts on mass 1 is to the left (negative), aiming to restore the spring to its equilibrium length.
2) The forces acting on mass [asciimath]m_2[/asciimath]are
- Restorative Force of the spring [asciimath]k_3[/asciimath] , [asciimath]F_(s1)=-k_3x_2[/asciimath],
- Restorative Force of the spring [asciimath]k_2[/asciimath] , [asciimath]F_(s2)=k_2(x_2-x_1)[/asciimath], where [asciimath]x_2-x_1[/asciimath] is the displacement of the middle spring.
- Damping Force [asciimath]F_(d1)=-c_2x_2'[/asciimath], where [asciimath]c_2[/asciimath] is the damping coefficient for the damper 2. If present, the damping force is proportional to the velocity of the mass and acting in the opposite direction of motion.
- External Force [asciimath]F_2(t)[/asciimath]. It includes any external force acting on mass [asciimath]m_2[/asciimath], which might be periodic or random, leading to forced vibrations.
According to Newton’s second law,
[asciimath]ma=sumF[/asciimath]
[asciimath]m_2x''_2=-c_2x'_2-k_3x_2-k_2(x_2-x_1)+F_2(t)[/asciimath]
This equation simplifies to
[asciimath]m_2x''_2=-c_2x'_2+k_2x_1 -(k_2+k_3)x_2+F_2(t)[/asciimath]
Note that since [asciimath]x_2ltx_1[/asciimath], the spring [asciimath]k_2[/asciimath] is compressed, and the force it exerts on mass 2 is to the right and should be positive, which is consistent with the sign of [asciimath]-k_2(x_2-x_1)>0[/asciimath].
Therefore, the time-dependent displacements of the masses are described by the system of differential equations
[asciimath]m_1x''_1=-c_1x'_1-(k_1+k_2)x_1+k_2x_2+F_1(t)[/asciimath]
[asciimath]m_2x''_2=-c_2x'_2+k_2x_1 -(k_2+k_3)x_2+F_2(t)[/asciimath]
a) Rewrite the derived system of differential equations in Example 6.10.2 to a system of first-order differential equations. b) Write the system in matrix form.
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a) The equations governing the mass-spring system in Figure 6.10.1 are derived in Example 6.10.2.
[asciimath]m_1x''_1=-c_1x'_1-(k_1+k_2)x_1+k_2x_2+F_1(t)[/asciimath]
[asciimath]m_2x''_2=-c_2x'_2+k_2x_1 -(k_2+k_3)x_2+F_2(t)[/asciimath]
Section 6.4-C discussed how to convert higher-order differential equations as a system of first-order equations. We introduce new variables as follows:
[asciimath]y_1=x_1[/asciimath] , [asciimath]y_2=x'_1[/asciimath] , [asciimath]y_3=x_2[/asciimath] , and [asciimath]y_4=x'_2[/asciimath]
The equations then can be written as
[asciimath]y'_1=y_2[/asciimath]
[asciimath]m_1y'_2=-c_1y_2-(k_1+k_2)y_1+k_2y_3+F_1(t)[/asciimath]
[asciimath]y'_3=y_4[/asciimath]
[asciimath]m_2y'_4=-c_2y_4+k_2y_1 -(k_2+k_3)y_3+F_2(t)[/asciimath]
b) The system in the matrix form is
[asciimath]bb(y)'(t)=Abb(y)(t)+bb(f)(t)[/asciimath]
[asciimath]bb(y)'(t)=[(0,1,0,0),(-(k_1+k_2)/(m_1),(c_1)/(m_1),(k_2)/(m_1),0),(0,0,0,1),((k_2)/(m_2),0,-(k_2+k_3)/(m_2) ,-(c_2)/(m_2))][/asciimath] [asciimath]bb(y)(t)+[(0),((F_1(t))/(m_1)),(0),((F_2(t))/(m_2)) ][/asciimath]
Consider a coupled mass-spring system, as described in Example 6.10.2, with the following parameters: both masses [asciimath]m_1[/asciimath] and [asciimath]m_2[/asciimath] are 1 kg, and all spring constants [asciimath]k_1[/asciimath], [asciimath]k_2[/asciimath], and [asciimath]k_3[/asciimath] are 1 N/m. The system is isolated from external forces and damping effects. Initially, the displacements and velocities are given as [asciimath]x_1(0)=0[/asciimath] m, [asciimath]x'_1(0)=0[/asciimath] m/s, [asciimath]x_2(0)=1[/asciimath] m, and [asciimath]x'_2(0)=0[/asciimath] m/s. Solve the initial value problem to determine the displacements of the masses as functions of time. Due to the complexity of calculations, use matrix algebra software to find the eigenvalues and eigenvectors.
Show/Hide Solution
Given information:
- [asciimath]m_1=m_2=1\ "kg"[/asciimath]
- [asciimath]k_1=k_2=k_3=1\ "N"//"m"[/asciimath]
- No damping: [asciimath]c_1=c_2=0[/asciimath]
- No External forces: [asciimath]F_1=F_2=0[/asciimath]
- Initial conditions:[asciimath]x_1(0)=0[/asciimath], [asciimath]x_2(0)=1\ "m"[/asciimath], [asciimath]x'_1(0)=x_2'(0)=0[/asciimath]
In Example 6.10.3, we converted a second-order system governing the coupled mass-spring system to a first-order system and expressed it in matrix notation.
[asciimath]bb(y)'(t)=Abb(y)(t)+bb(f)(t)[/asciimath]
[asciimath]bb(y)'(t)=[(0,1,0,0),(-(k_1+k_2)/(m_1),(c_1)/(m_1),(k_2)/(m_1),0),(0,0,0,1),((k_2)/(m_2),0,-(k_2+k_3)/(m_2) ,-(c_2)/(m_2))][/asciimath] [asciimath]bb(y)(t)+[(0),((F_1(t))/(m_1)),(0),((F_2(t))/(m_2)) ][/asciimath]
where
[asciimath]bb(y)=[(y_1),(y_2),(y_3),(y_4)]=[(x_1),(x'_1),(x_2),(x'_2)][/asciimath]
Substituting the given values, the initial value problem becomes
[asciimath]bb(y)'(t)=[(0,1,0,0),(-2,0,1,0),(0,0,0,1),(1,0,-2 ,0)][/asciimath] [asciimath]bb(y)(t), \ \ bb(y)(0)=[(0),(0),(1),(0) ][/asciimath]
Using matrix algebra software, we find the eigenvalues and eigenvectors of the coefficient matrix. The eigenvalues and eigenvectors occur in the following complex conjugate pairs.
Eigenvector for [asciimath]lambda_(1,2)=+-i[/asciimath] : [asciimath]bb(u)_(1,2)=[(0),(1),(0),(1)]+-i [(-1),(0),(-1),(0)][/asciimath]
Eigenvector for [asciimath]lambda_(3,4)=+-sqrt(3)i[/asciimath] : [asciimath]bb(u)_(3,4)=[(0),(-3),(0),(3)]+-i [(sqrt(3)),(0),(-sqrt(3)),(0)][/asciimath]
The solution for each pair is given by Equation 6.7.1.
For the first pair, [asciimath]lambda_1,2=+-i[/asciimath], the two linearly independent solutions are
[asciimath]bb"y"_1=[(0),(1),(0),(1)] cos(t)-[(-1),(0),(-1),(0)] sin( t)=[/asciimath] [asciimath][(sin(t)),(cos(t)),(sin(t)),(cos(t))][/asciimath]
[asciimath]bb"y"_2=[(0),(1),(0),(1)] sin(t)+[(-1),(0),(-1),(0)]cos( t)=[/asciimath] [asciimath][(-cos(t)),(sin(t)),(-cos(t)),(sin(t))][/asciimath]
For the second pair, [asciimath]lambda_3,4=+-sqrt(3)i[/asciimath], the two linearly independent solutions are
[asciimath]bb"y"_3=[(0),(-3),(0),(3)] cos(sqrt(3) t)-[(sqrt(3)),(0),(-sqrt(3)),(0)] sin( sqrt(3) t)=[/asciimath] [asciimath][(-sqrt(3) sin(sqrt(3) t)),(-3cos(sqrt(3) t)),(sqrt(3) sin(sqrt(3) t)),(3cos(sqrt(3) t))][/asciimath]
[asciimath]bb"y"_4=[(0),(-3),(0),(3)] sin(sqrt(3) t)+[(sqrt(3)),(0),(-sqrt(3)),(0)] cos( sqrt(3) t)=[/asciimath] [asciimath][(sqrt(3) cos(sqrt(3) t)),(-3sin(sqrt(3) t)),(-sqrt(3) cos(sqrt(3) t)),(3sin(sqrt(3) t))][/asciimath]
Thus, the fundamental matrix of the system is
[asciimath]Y(t)=[/asciimath] [asciimath][(bb(y)_1,bb(y)_2, bb(y)_3, bb(y)_4)][/asciimath]
[asciimath]= [(sin(t),-cos(t),-sqrt(3) sin(sqrt(3) t),sqrt(3) cos(sqrt(3) t)),(cos(t),sin(t),-3 cos(sqrt(3) t),-3sin(sqrt(3) t)),(sin(t),-cos(t),sqrt(3) sin(sqrt(3) t),-sqrt(3) cos(sqrt(3) t)),(cos(t),sin(t),3 cos(sqrt(3) t),3 sin(sqrt(3) t))][/asciimath]
The general solution to the system in matrix form is
[asciimath]bb(y)(t)= [(sin(t),-cos(t),-sqrt(3) sin(sqrt(3) t),sqrt(3) cos(sqrt(3) t)),(cos(t),sin(t),-3 cos(sqrt(3) t),-3sin(sqrt(3) t)),(sin(t),-cos(t),sqrt(3) sin(sqrt(3) t),-sqrt(3) cos(sqrt(3) t)),(cos(t),sin(t),3 cos(sqrt(3) t),3 sin(sqrt(3) t))][/asciimath] [asciimath][(c_1),(c_2),(c_3),(c_4)][/asciimath]
Applying the initial conditions, we obtain
[asciimath]bb(y)(0)=[(0),(0),(1),(0) ][/asciimath]
[asciimath][(0,-1,0,sqrt(3)),(1,0,-3,0),(0,-1,0,-sqrt(3)),(1,0,3 ,0)][/asciimath] [asciimath][(c_1),(c_2),(c_3),(c_4)][/asciimath] [asciimath]=[(0),(0),(1),(0) ][/asciimath]
Solving for the coefficients, we find
[asciimath]c_1=0,\ c_2=-1/2,\ c_3=0,\ c_4=-sqrt3/6[/asciimath]
Therefore, the solution to the initial value problem is
[asciimath]bb(y)(t)= [(sin(t),-cos(t),-sqrt(3) sin(sqrt(3) t),sqrt(3) cos(sqrt(3) t)),(cos(t),sin(t),-3 cos(sqrt(3) t),-3sin(sqrt(3) t)),(sin(t),-cos(t),sqrt(3) sin(sqrt(3) t),-sqrt(3) cos(sqrt(3) t)),(cos(t),sin(t),3 cos(sqrt(3) t),3 sin(sqrt(3) t))][/asciimath] [asciimath][(0),(-1/2),(0),(-sqrt(3)/6)][/asciimath]
This can be written as
[asciimath]bb(y)=[(y_1),(y_2),(y_3),(y_4)][/asciimath] [asciimath]=[/asciimath] [asciimath][(1/2cos(t)-1/2cos(sqrt(3)t)),(-1/2sin(t)+(sqrt(3))/2sin(sqrt(3)t)),(1/2cos(t)+1/2cos(sqrt(3)t)),(-1/2sin(t)-(sqrt(3))/2sin(sqrt(3)t))][/asciimath]
Recall from Example 10.6.3 that we introduced variables [asciimath]bb(y)_n[/asciimath] to represent the displacements and velocities of the masses.
[asciimath][(y_1),(y_2),(y_3),(y_4)]=[(x_1),(x'_1),(x_2),(x'_2)][/asciimath]
Given this conversion, the displacements of mass 1 ([asciimath]x_1[/asciimath]) and mass 2 ([asciimath]x_2[/asciimath]) as a function of time are determined by
[asciimath]x_1=y_1=1/2cos(t)-1/2cos(sqrt(3)t)[/asciimath]
[asciimath]x_2=y_3=1/2cos(t)+1/2cos(sqrt(3)t)[/asciimath]
The below visualization displays the displacements of the two masses over time in a coupled mass-spring system in this example, plotted on a graph with time on the horizontal axis (ranging from 0 to 10 seconds) and displacement in meters on the vertical axis. The line for Mass 1 oscillates, indicating a pattern of motion that varies between positive and negative displacements, suggesting complex harmonic motion. The line for Mass 2 follows a similar oscillatory pattern, but with phase and amplitude differences compared to Mass 1, reflecting the interaction between the two masses through the spring system.
Section 6.10 Exercises
- Consider a coupled mass-spring system, as described in Example 6.10.2, with the following parameters: both masses [asciimath]m_1[/asciimath] and [asciimath]m_2[/asciimath] are 1 kg, and all spring constants [asciimath]k_1[/asciimath], [asciimath]k_2[/asciimath], and [asciimath]k_3[/asciimath] are 1 N/m. The system is isolated from external forces and damping effects. Initially, the displacements and velocities are given as [asciimath]x_1(0)=0[/asciimath] m, [asciimath]x'_1(0)=2[/asciimath] m/s, [asciimath]x_2(0)=0[/asciimath] m, and [asciimath]x'_2(0)=0[/asciimath] m/s. Solve the initial value problem to determine the displacements of the masses as functions of time. Use matrix algebra software to find the eigenvalues and eigenvectors.
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[asciimath]x_1=sin(t)+sqrt(3)/3sin(sqrt(3)t)[/asciimath]
[asciimath]x_2=sin(t)-sqrt(3)/3sin(sqrt(3)t)[/asciimath]
